# Queuing Theory with Excel - extra

Only a few days ago, I blithly said

The Excel formula of 1/(Poisson($c$,$c\rho$,false)*exp($c\rho$)/($1-\rho$) + Poisson($c$,$c\rho$,true)*exp($c\rho$) - Poisson($c$,$c\rho$,false)*exp($c\rho$))

Is equal to
$\biggl [1 + \frac{(c\rho)^c}{c!(1-\rho)} + \sum\limits_{n=1}^{c-1} \frac{(c\rho)^n}{n!} \biggr]^{-1}=\wp_0$

Now you have the "tricky" $\wp_o$ value, all other values in the M/M/c queuing theory are MUCH easier to calculate.

It has been suggested to me, that it might help if I made this formula a little easier still.

Never one to shirk a challenge.....

What can we do with this?
Starting with the formula

1/(Poisson($c$,$c\rho$,false)*exp($c\rho$)/($1-\rho$) + Poisson($c$,$c\rho$,true)*exp($c\rho$) - Poisson($c$,$c\rho$,false)*exp($c\rho$))

The exp($c\rho$) gets used quite a lot in that formula, so we can take it out and use it only once.
1/(exp($c\rho$) * (Poisson($c$,$c\rho$,false)/($1-\rho$) + Poisson($c$,$c\rho$,true) - Poisson($c$,$c\rho$,false)))

Already, that looks much simpler to the eye.

Or if you want to limit the number of times you have to use the Poisson formula, you could then put those two Poisson($c$,$c\rho$,false) statements together and say
1/(exp($c\rho$) * (Poisson($c$,$c\rho$,true) + (Poisson($c$,$c\rho$,false)*($\frac{\rho}{1-\rho}$) )))

Checking that you've got it right
Of course. It is always useful to check that you have entered formula correctly. For this purpose, I always try a "simple" calculation.

Let's look at an example queue in which there are 5 servers ($c$), events occur every 6 minutes and the service time is 20 minutes.
The arrival rate is $\lambda = \frac{1}{6}$
The service rate is $\mu = \frac{1}{20}$
So the Utilisation $U$ or $\rho$ is $\frac{\lambda}{c\mu}$ which is $\frac{0.167}{5 * 0.05}$ = 0.6667

$\biggl [1 + \frac{(c\rho)^c}{c!(1-\rho)} + \sum\limits_{n=1}^{c-1} \frac{(c\rho)^n}{n!} \biggr]^{-1}$

$\biggl [1 + \frac{(5*0.6667)^5}{5!(1-0.6667)} + \sum\limits_{n=1}^{4} \frac{(5*0.6667)^n}{n!} \biggr]^{-1}$

$\biggl [1 + 10.28806 + (3.3333 + 5.555555 + 6.17284 + 5.14403) \biggr]^{-1} = 0.03175$

Excel Formula
1 / (exp($c\rho$) * (Poisson($c$,$c\rho$,true) + (Poisson($c$,$c\rho$,false) * ($\frac{\rho}{1-\rho}$) )))

1 / (EXP(5*0.6667) * (POISSON(5,5*0.6667,TRUE) + (POISSON(5,5*0.6667,FALSE) * (0.6667/(1-0.6667)))))

1 / (28.0363 * (0.8788 + (0.1223 * (2) ) ) ) = 0.03175

So. I now know that I have got the right Excel formula (with the right brackets in the right place). I can now continue with my queuing model, fully able to calculate queue probability in a single cell, even for very large systems with a large number of servers.

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